`4y^2-y-18`=$(2y)^2-2.2y.\frac{1}{4} +\frac{1}{16}-\frac{1}{16}-18$
=$(2y-\frac{1}{4})^2-\frac{1}{16}-\frac{288}{16}$
=$(2y-\frac{1}{4})^2-\frac{289}{16}$
=$(2y-\frac{1}{4})^2-(\frac{17}{4})^2$
=$(2y-\frac{1}{4}-\frac{17}{4})(2y-\frac{1}{4}+\frac{17}{4})$
=$(2y-\frac{18}{4})(2y+\frac{16}{4})$
=$(2y-\frac{9}{2})(2y+4)$